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RMS value of a fully rectified clamped sinusoid.

  1. Apr 18, 2013 #1
    so i have a question about calculating the RMS value of a fully rectified clamped sinusoid.

    Assumptions:
    The top of the waveform = U
    It is clamped at 0.5 U

    I can calculate the RMS value by adding the 3 components of the wave, ei. @ 0.5U ω = π/6 & 5π/6 which forms a block and two side compnents. I then calculate integral of the squares of the two side components of the sinusoid and the block wave spanned from π/6 to 5π/6. and i find the RMS value which is ≈ 0.44*U.

    Now i tried to calculate this a different way, but i could not get it to work. I thought it was logically equivalent to adding the components, but it is not working. Maybe someone could shed some light on my fallacies.
    What i am trying to do is to calculate the RMS value of the complete rectified sinusoid and then substract the RMS value of the piece of sinusoid that has been clamped off. I found the piece that has been clamped off to be:

    √(1/π * ∫(from π/6 to 5π/6) (-0.5*U + U*sin(ωt))^2 dωt) {-0.5*U drops the sinusoid so that i can find the RMS value of only the clamped of part. At least this is what i assume...}

    and so i subtract that from the RMS value of the fully rectified sinusoid which is U/√2

    This however provides me with a wrong answer which was somewhere around 0.27*U . What am i wrongfully assuming or calculating?
     
  2. jcsd
  3. Apr 20, 2013 #2

    Stephen Tashi

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    Science Advisor

    I haven't read all the details of your work, but In general, [itex] \int {(f(x) + g(x))^2 dx} [/itex] is not equal to [itex] \int {f^2(x) dx} + \int {g^2(x) dx} [/itex]. So you can't compute the RMS by expressing a wave as a sum of arbitrary components. You must choose the components to be orthogonal functions .
     
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